Ta có : \(\left|2x-3\right|+12=2\)
TH1 : \(2x-3\ge0\left(x\ge\frac{3}{2}\right)\)
=> \(\left|2x-3\right|=2x-3\)
Nên ta có phương trình : \(2x-3+12=2\)
=> \(2x=-7\)
=> \(x=-\frac{7}{2}\left(KTM\right)\)
TH2 : \(2x-3< 0\left(x< \frac{3}{2}\right)\)
=> \(\left|2x-3\right|=3-2x\)
Nên ta có phương trình : \(3-2x+12=2\)
=> \(-2x=-13\)
=> \(x=\frac{13}{2}\left(KTM\right)\)
Vậy phương trình vô nghiệm .
\(\left|2x-3\right|+12=2\)
\(TH1:2x-3+12=2\)
\(2x-3=2-12\)
\(2x-3=-10\)
\(2x\) \(=-10+3\)
\(2x\) \(=-7\)
\(x\) \(=-7\div2\)
\(x\) \(=\frac{-7}{2}\)
\(TH2:2x-3+12=-2\)
\(2x-3=-2-12\)
\(2x-3=-14\)
\(2x\) \(=-14+3\)
\(2x\) \(=-9\)
\(x\) \(=-9\div2\)
\(x\) \(=\frac{-9}{2}\)
Vậy \(x\left\{\frac{-7}{2};\frac{-9}{2}\right\}\)
